Prove that $\cos\frac{1}{\sqrt{2019}}$ is irrational.
Would someone give me a hand on this question? Even a hint is appreciated. I don't have any clue.
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Perhaps, you could use the Taylor series $\cos x = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k)!}.$ – Dylan C. Beck May 21 '20 at 16:01
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From here, a naïve guess would be to assume that the given number is rational, and derive a contradiction. If you need some inspiration, you might read this Wikipedia entry: https://en.wikipedia.org/wiki/Proof_that_e_is_irrational. – Dylan C. Beck May 21 '20 at 16:03
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@Carlo, I have tried it, but it seems a dead end to me. – yangcs11 May 21 '20 at 16:06
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1This follows from the Lindemann-Weierstrass Theorem – lulu May 21 '20 at 16:08
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@Carlo, Thanks a lot! – yangcs11 May 21 '20 at 16:09
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$2019$ is square free and may be written as the product of $45 \pm \sqrt{6}$. Are these two numbers irrational? What happens when you take their square root? Can you then form partial fractions and use Trig sum angle formulae? – May 21 '20 at 16:14
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Neither prime numbers nor rational numbers come into play here. – Peter May 22 '20 at 07:42
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1See the answer of this question. – May 22 '20 at 10:50
2 Answers
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Following the suggestions of @Carlo, I solved this problem myself. It is indeed similar to the proof that $e$ is irrational. Thinking it might help someone else, so I just post it in the following.
Suppose it is rational number, such that $$\cos\frac{1}{\sqrt{2019}} = p/q,$$
where $p$ and $q$ are co-primes. Expand the cosine as sum of power series, which is $$\cos\frac{1}{\sqrt{2019}} = \sum_{i=0}^\infty \frac{(-)^i}{(2i)!}\frac{1}{2019^i}.$$
Define the sum $$S_n = \sum_{i=0}^nx_i,\quad \mathrm{and}\quad x_i=\frac{(-)^i}{(2i)!}\frac{1}{2019^i}$$
It is reasonable to assume for some integer large enough $N\gg1$, we have $$x_{2m}+x_{2m+1}>0,\ \forall\ m\geqslant N,$$ also $$x_{2m+1}+x_{2m+2}<0,\ \forall\ m\geqslant N,$$
So it is easy to see that $$0<p/q-S_{2N-1}<x_{2N}=\frac{1}{(4N)!2019^{2N}}.$$
Thus if we choose, for example, that $4N>q$, we will have $$0<(4N)!2019^{2N}\times(p/q-S_{2N-1})<1.$$
But on the other hand, $(4N)!2019^{2N}\times(p/q-S_{2N-1})$ must be an integer. Thus by contradiction, it must be irrational.
yangcs11
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Yes. A little simpler: If $x=\cos 1/\sqrt {2019}\in \Bbb Q$ then $(2n)!x,$ and, a fortiori, $(2019)^n(2n)!x,$ are integers for any sufficiently large $n,$ but that never happens. – DanielWainfleet May 23 '20 at 06:52
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If $x\in \Bbb R$ and for every $r>0$ there exists $<a,b>\in \Bbb Z\times \Bbb Z^+$ such that $0<|x-a/b|<r/b,$ then $x \not \in \Bbb Q.$
Proof: If $x=c/d$ with $c\in \Bbb Z$ and $d\in \Bbb Z^+$ then for any $<a,b>\in \Bbb Z\times \Bbb Z^+$ we have $$0<|x-a/b|\implies 0<|c/d-a/b|\implies 0<|cb-ad|\implies$$ $$\implies 1\le |cd-ab|\implies (1/d)/b\le (1/bd)|cb-ad|=|x-a/b|.$$ (I.e. if $x=c/d$ then the "$r$" of my 1st sentence cannot be less than $1/d$.)
Apply this with $n\in \Bbb Z^+$ large enough that $r<(2n)!(2019)^n\ \cdot|\sum_{j=n+1}^{\infty}(-1)^j/((2j)!2019^j)\,|$
with $b=(2n)!(2019)^n$ and $a/b=\sum_{j=0}^n (-1)^j/((2j)!2019^j).$
DanielWainfleet
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Thank you for pointing out this theorem! It really simplifies the proof. – yangcs11 May 23 '20 at 13:28
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A common manoevre in the proof: Whenever you have an inequality $p<q$ between integers, it can be "sharpened" to $ p+1\le q.$ – DanielWainfleet May 23 '20 at 18:15